package com.dycong.common.leetcode;

import com.dycong.common.leetcode.ds.ListNode;

/**
 * 将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 
 * <p>
 * 示例：
 * <p>
 * 输入：1->2->4, 1->3->4
 * 输出：1->1->2->3->4->4
 * <p>
 * 来源：力扣（LeetCode）
 * 链接：https://leetcode-cn.com/problems/merge-two-sorted-lists
 * 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
 *
 * @author dycong
 * @date 2019/12/2 12:35
 */
public class MergeTwoLists_21 {

    public static void main(String[] args) {
        MergeTwoLists_21 mergeTwoLists_21 = new MergeTwoLists_21();

    }

    /**
     * 遍历
     *
     * @param l1
     * @param l2
     * @return
     */
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {

        if (null == l1) return l2;
        if (null == l2) return l1;

        ListNode head, curNode;

        if (l1.val > l2.val) {
            head = curNode = l2;
            l2 = l2.next;
        } else {
            head = curNode = l1;
            l1 = l1.next;
        }

        while (l1 != null || l2 != null) {
            ListNode node;
            if (l1 == null) {
                node = l2;
                l2 = l2.next;
            } else if (l2 == null) {
                node = l1;
                l1 = l1.next;
            } else {
                if (l1.val > l2.val) {
                    node = l2;
                    l2 = l2.next;
                } else {
                    node = l1;
                    l1 = l1.next;
                }
            }
            curNode.next = node;
            curNode = curNode.next;
        }
        return head;
    }

    /**
     * 递归
     *
     * @param l1
     * @param l2
     * @return
     */
    public ListNode mergeTwoLists_(ListNode l1, ListNode l2) {
        if (null == l1)
            return l2;
        if (null == l2)
            return l1;
        if (l1.val > l2.val) {
            l2.next = mergeTwoLists_(l1, l2.next);
            return l2;
        } else {
            l1.next = mergeTwoLists_(l1.next, l2);
            return l1;
        }
    }
}
